Topology

Topological spaces, compactness and connectedness

A topology is the minimal structure needed to talk about continuity, open sets, compactness and connectedness without any notion of distance; continuous maps send compact sets to compact sets and connected sets to connected sets, the two facts behind the general Intermediate Value Theorem, robot motion planning, and existence proofs in economics.

IntuitionWhat survives without a ruler

Stretch or bend a rubber sheet and distances change completely, but some facts never change: a loop around a hole stays a loop around a hole, a single piece stays a single piece, and a solid disk never tears into two. A topology captures exactly this "rubber-sheet" structure by remembering only which sets count as open, throwing away distance altogether.

A torus: rotate it and stretch it and it stays the same topological space. The loop that goes around the central hole can never be shrunk to a point while staying on the surface — a fact that survives every continuous deformation.

UndergraduateThe axioms of a topology

Definition: Topological space

A topology on a set $X$ is a collection $\tau$ of subsets of $X$ (called open sets) satisfying: (T1) $\emptyset \in \tau$ and $X \in \tau$ ; (T2) the union of any collection of sets in $\tau$ (even infinitely many) is in $\tau$ ; (T3) the intersection of any finite collection of sets in $\tau$ is in $\tau$ . The pair $(X,\tau)$ is a topological space. Every metric space is a topological space: declare $U$ open exactly when every point of $U$ has an open ball around it that stays inside $U$ .

\emptyset, X \in \tau \qquad \bigcup_{i \in I} U_i \in \tau \ (U_i \in \tau) \qquad \bigcap_{i=1}^n U_i \in \tau \ (U_i \in \tau)

With open sets as the only primitive, continuity is redefined without any $\varepsilon$ - $\delta$ : a function $f: X \to Y$ between topological spaces is continuous exactly when the preimage of every open set is open, $f^{-1}(U)$ open in $X$ for every $U$ open in $Y$ . On metric spaces this matches the familiar $\varepsilon$ - $\delta$ definition exactly, but it now also makes sense on spaces with no distance at all.

f: X \to Y \text{ continuous} \iff f^{-1}(U) \in \tau_X \ \ \forall U \in \tau_Y

A topological space is Hausdorff if any two distinct points can be separated by disjoint open sets: for all $x \ne y$ there exist $U, V \in \tau$ with $x \in U$ , $y \in V$ , and $U \cap V = \emptyset$ . Every metric space is automatically Hausdorff (take balls of radius $d(x,y)/2$ ), but strange topologies exist without this property, where limits of sequences are not even unique.

\forall x \ne y\ \exists U, V \in \tau:\ x \in U,\ y \in V,\ U \cap V = \emptyset

Compactness across settings
Setting	Compactness test	Example
$\mathbb{R}^n$ (Heine–Borel)	closed and bounded	$[0,1]$ is compact, $(0,1)$ is not
general metric space	sequential compactness $=$ open-cover compactness	every sequence has a convergent subsequence
general topological space	only open-cover compactness applies	every open cover has a finite subcover

UndergraduateTheorems

Theorem: Continuous images of compact sets are compact

If $f: X \to Y$ is continuous and $K \subseteq X$ is compact (every open cover of $K$ has a finite subcover), then $f(K) \subseteq Y$ is compact. Consequently, if $Y$ is Hausdorff, every compact subset $K \subseteq Y$ is closed.

Why is it true?

It explains why continuous functions on closed bounded intervals always attain a maximum, and it is the abstract engine behind "compact + Hausdorff = as good as a closed bounded set" throughout analysis.

Proof

Step 1 — Pull back an arbitrary open cover of $f(K)$ . Let $\{V_i\}_{i \in I}$ be any collection of open sets in $Y$ covering $f(K)$ , i.e. $f(K) \subseteq \bigcup_i V_i$ . Since $f$ is continuous, each $f^{-1}(V_i)$ is open in $X$ . For every $x \in K$ , $f(x) \in f(K)$ lies in some $V_i$ , so $x \in f^{-1}(V_i)$ ; thus $\{f^{-1}(V_i)\}_{i \in I}$ is an open cover of $K$ .

Step 2 — Use compactness of $K$ to extract a finite subcover. Since $K$ is compact, finitely many of these sets already cover $K$ : there exist $i_1, \dots, i_n$ with $K \subseteq f^{-1}(V_{i_1}) \cup \dots \cup f^{-1}(V_{i_n})$ .

Step 3 — Push the finite subcover forward. Applying $f$ to both sides, $f(K) \subseteq f\big(f^{-1}(V_{i_1}) \cup \dots \cup f^{-1}(V_{i_n})\big) \subseteq V_{i_1} \cup \dots \cup V_{i_n}$ (using $f(f^{-1}(V)) \subseteq V$ always). So $\{V_{i_1}, \dots, V_{i_n}\}$ is a finite subcover of $f(K)$ drawn from the original cover. Since $\{V_i\}$ was arbitrary, $f(K)$ is compact.

Step 4 — The corollary: compact subsets of Hausdorff spaces are closed. Let $K \subseteq Y$ be compact, $Y$ Hausdorff, and fix any $y \notin K$ . For each $x \in K$ , Hausdorff separation gives disjoint open sets $U_x \ni x$ and $W_x \ni y$ . The sets $\{U_x\}_{x \in K}$ cover $K$ , so finitely many $U_{x_1}, \dots, U_{x_n}$ already cover $K$ by compactness. Then $W = W_{x_1} \cap \dots \cap W_{x_n}$ is a finite intersection of open sets (hence open), contains $y$ , and is disjoint from every $U_{x_j}$ , hence disjoint from $K \subseteq \bigcup_j U_{x_j}$ . So every point outside $K$ has an open neighborhood missing $K$ : the complement of $K$ is open, i.e. $K$ is closed.

Theorem: Continuous images of connected spaces are connected (general Intermediate Value Theorem)

If $f: X \to Y$ is continuous and $X$ is connected (it cannot be written as a union of two disjoint nonempty open sets), then $f(X)$ is connected. In particular, if $X$ is connected and $f: X \to \mathbb{R}$ is continuous, then $f(X)$ is an interval: for any $a, b \in X$ , $f$ attains every value between $f(a)$ and $f(b)$ .

Why is it true?

This is the true source of the Intermediate Value Theorem from calculus — connectedness, not the specific formula of $f$ , is what forces every intermediate value to be hit, and the same argument works for any connected domain, not just intervals of $\mathbb{R}$ .

Proof

Step 1 — Suppose for contradiction that $f(X)$ is disconnected. Then $f(X) = A \cup B$ for some disjoint, nonempty sets $A, B$ that are each open in the subspace topology of $f(X)$ : there exist open sets $A', B'$ in $Y$ with $A = f(X) \cap A'$ and $B = f(X) \cap B'$ .

Step 2 — Pull the separation back to $X$ . Let $P = f^{-1}(A')$ and $Q = f^{-1}(B')$ . Since $f$ is continuous, $P$ and $Q$ are open in $X$ . Every $x \in X$ has $f(x) \in f(X) = A \cup B$ , so $f(x) \in A'$ or $f(x) \in B'$ , meaning $x \in P$ or $x \in Q$ : thus $X = P \cup Q$ . Also $P$ and $Q$ are both nonempty (since $A, B$ are nonempty and are hit by $f$ ), and $P \cap Q = \emptyset$ (if $x \in P \cap Q$ then $f(x) \in A' \cap B' \cap f(X) = A \cap B = \emptyset$ , impossible).

Step 3 — Contradiction. $P$ and $Q$ are disjoint nonempty open sets with $X = P \cup Q$ , exactly the definition of $X$ being disconnected. This contradicts the hypothesis that $X$ is connected. So $f(X)$ cannot be disconnected: it is connected.

Step 4 — The interval corollary. The connected subsets of $\mathbb{R}$ are exactly the intervals (a standard fact: any subset that skips a real number between two of its points fails connectedness via the same open-set separation). Since $f(X)$ is connected by Steps 1–3, $f(X)$ is an interval of $\mathbb{R}$ , so it contains every real number between any two of its elements $f(a)$ and $f(b)$ — the classical Intermediate Value Theorem, now seen as a special case of a purely topological fact.

UndergraduateReal-World Applications and Worked Examples

A robot arm's configuration space (all joint-angle combinations) is a topological space; whether the robot can move from pose $a$ to pose $b$ without collision is exactly whether $a$ and $b$ lie in the same connected component of the obstacle-free region, and boundedness/compactness of joint limits guarantees the reachable set behaves well (is closed, attains extremes). In economics, Arrow–Debreu equilibrium existence proofs rely on continuity of excess-demand functions on a compact price simplex, combined with connectedness arguments (a 1-D special case being: if excess demand is positive at one price and negative at another, continuity and connectedness of the price interval force an equilibrium price where it is exactly zero).

Example: Is the free configuration space connected?

A 2-link robot arm has configuration space $T^2 = S^1 \times S^1$ (angles $\theta_1, \theta_2$ each range over a circle). An obstacle forbids the single point $(\theta_1,\theta_2) = (0,0)$ . Is the free configuration space $T^2 \setminus \{(0,0)\}$ still connected, i.e. can the arm reach any pose from any other without hitting the obstacle?

Solution

Step 1 — Recall that removing a single point from a connected space of dimension $\ge 2$ typically preserves connectedness (unlike dimension 1, where removing a point from an interval disconnects it). The torus $T^2$ is a connected 2-dimensional surface.

Step 2 — Construct an explicit path avoiding the obstacle. Given any two poses $p, q \in T^2 \setminus \{(0,0)\}$ , pick a path from $p$ to $q$ along the torus that happens to pass exactly through $(0,0)$ (such a path always exists since $T^2$ is path-connected). If it does pass through the obstacle, perturb the path slightly near that instant, routing it around $(0,0)$ through a small detour on the surface (possible since $T^2$ minus a point still has "room" in the second dimension to go around).

Step 3 — Conclude connectedness. Since any two points can always be joined by a path avoiding the single removed point, $T^2 \setminus \{(0,0)\}$ is path-connected, hence connected: the robot can indeed reach any pose from any other despite the single-point obstacle. (This is genuinely different from a 1-link arm, where the configuration space is just $S^1$ , and removing a single point would disconnect the reachable region into a single arc — no detour is possible in one dimension.)

Example: Finding a 1-good market equilibrium price by connectedness

A market's excess demand function $g(p)$ (demand minus supply at price $p$ ) is continuous on the price interval $[1,10]$ , with $g(1) = 50 > 0$ (shortage at low price) and $g(10) = -20 < 0$ (surplus at high price). Explain, using the connectedness-based theorem, why an equilibrium price $p^* \in [1,10]$ with $g(p^*)=0$ must exist, without assuming any specific formula for $g$ .

Solution

Step 1 — Identify the topological ingredients. $[1,10]$ is a connected subset of $\mathbb{R}$ (an interval), and $g: [1,10] \to \mathbb{R}$ is continuous by assumption.

Step 2 — Apply the connectedness theorem. By the theorem, $g([1,10])$ is a connected subset of $\mathbb{R}$ , hence an interval. Since $g(1)=50$ and $g(10)=-20$ both lie in $g([1,10])$ , the interval $g([1,10])$ must contain every real number between $-20$ and $50$ — in particular it must contain $0$ .

Step 3 — Conclude existence (not uniqueness) of equilibrium. Therefore there exists at least one $p^* \in [1,10]$ with $g(p^*) = 0$ : an equilibrium price exists. Crucially, this argument used only continuity and connectedness — no formula for $g$ , no calculus, no convexity assumption — showing why the topological approach generalizes so well to economic models where demand curves are not given by closed-form formulas.

Which of these is NOT one of the topology axioms?

A robot's free configuration space is disconnected into two separate pieces around obstacle. What does this mean physically?

An excess-demand function is continuous on the compact, connected price interval $[2,8]$ , positive at $p=2$ and negative at $p=8$ . What can you conclude?

$K \subseteq \mathbb{R}^2$ is compact and $f: \mathbb{R}^2 \to \mathbb{R}^2$ is continuous. What must be true of $f(K)$ ?

References

James Munkres (2000). Topology
John L. Kelley (1955). General Topology
Michael Farber (2008). Topology and Robot Motion Planning (survey chapter)